Problem: Multiply the following complex numbers: $({-4+i}) \cdot ({4+i})$
Solution: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({-4+i}) \cdot ({4+i}) = $ $ ({-4} \cdot {4}) + ({-4} \cdot {1}i) + ({1}i \cdot {4}) + ({1}i \cdot {1}i) $ Then simplify the terms: $ (-16) + (-4i) + (4i) + (1 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ -16 + (-4 + 4)i + 1i^2 $ After we plug in $i^2 = -1$ , the result becomes $ -16 + (-4 + 4)i - 1 $ The result is simplified: $ (-16 - 1) + (0i) = -17 $